Hi, I posted a thread a day or two back but still haven't managed to get any further with it :( to top it off I tried to move on to the next question and proceeded to get it and the next incorrect! HELLLP! Lol

Here are the questions I'm really struggling to cope with (disappointed as I thought I had this topic under wraps - apparently not!) Hoping that someone can offer me guidance on how I should tackle each of the questions... I've tried several methods of solving them and am still getting 0% :( I'll post what I have done so far for each of the questions so it may be easier to highlight where or what I am going/doing wrong.
Thank you in advance again to anyone who can help me on this, I really do appreciate it greatly.

Q1: A man deposits £1000 every half year into a pension fund earning 6% interest per annum, compunded half yearly. Deposits are made at the start of each half year. How much will the fund be worth at the end of 7.5 years after the initial deposit?

Q1 Working S7.5= 1000(1+6/200)^15 + 1000(1+6/200)^14 + 1000(1+6/200)^13 +... + 1000(1+6/200)^1
a=1.03, n=15, r=1.03
S15=1.03[(1-1.03^15)/(1-1.03)]
S15= 1000*1.03*18.6 = 19158


Q2: An investment account offers an interest rate of 8% per annum, compounded yearly. Assuming a single deposit of £39700 is placed in this investment account, how long would it take to amount to £68900 assuming that £100 is withdrawn at the end of the first two years and thereafter no further withdrawals are made?

Q2 Working
A1=39700*1.08-100= 42776
A2=42776*1.17-100=49947.92

Reinvest £49947.92 for t-2 years to get £68900.
68900=49947.9*1.08^t-2
log(68900/49947.92) = (t-2) log1.08
t-2=log(68900/49947.92)/log1.08 = 4.18
t= 6.18

Q3: An investment account offers an interest rate of 6% per annum compunded monthly. Assuming a single deposit of money is placed in this account, how much money must be invested now in order for the account to contain £45000 after 10 years?

Q3 Working: At=A0(1+r/100m)^mt

45000=Ao(1+6/1200)^1200

45000=Ao*1.005^1200

45000/1.005^1200=Ao

Ao=113.22

I already answered the first one on your other thread. (I couldn't come here for the past few days because of a phoneline problem cutting the internet)

2. I'm assuming that the withdrawal is done only once, 2 years after the deposit was made. (I would think about twice if they worded it that way: "assuming that £100 is withdrawn at the end of each of the first two years"

First, the manual way, I get it in between 7 and 8 years.

After year 1; 39700 x 1.08 = 42876
After year 2; 42876 x 1.08 = 46306.08

Withdrawal is made, 46306.08 - 100 = 46206.08

From there on, Money = 46306.08(1.08)^n

68900 = 46306.08(1.08)^n

1.4879... = 1.08^n

n = 5.16\ years

Hence, the total number of years = 5.16 + 2 = 7.16

(I added 2 years because of the two years not taken into account before the principal becomes 46306.08.)

3. Check again the value of mt. :p

Thanks for the reply as always! I used your formulae to compare my revised working for Q2 and although the working differed we ended up with the same answer (7 years).
Also your advise for Q3 was excellent as ever! (I really am dense at times lol)